wanminliu@gmail.com
Vector
calculus
cure
É
It
☐
✗
É
=/
I
J
E
¥
¥
¥
,
a
formal
cross
pr oduct
→
I
=p
,
⇐
g)
§
17
.
1
Gradient
,
Divergence
a'
formal
determinant
"
and
curl
Remand
For
two
vectors
a-
and
I
of
the
same
f-
:
IR
}
→
IR
f
=
f-
IX.
Y
.
Z
)
dimension
,
we
can
define
the
inner
product
☐
=
.gg
,
Zz
)
Differential
operat or
vet
"
a-
.
I
=
tail
151
•
'
'
ñab÷geg
a-
•
b-
Then
of
.
I
=
b-
•
a-
.
grad
f-
¥
Of
=
Cfx
.
fy
,
fz
)
But
here
we
borrow
the
notations
of
"
inner
pr oduct
"
a
vector
function
here
fx
=
¥
,
fy=
¥
fz=¥
it
is
not
a
really
inner
product
,
>
s
so
☐
•
=P
☒
É
•
P
f-
:
IR
>
→
IR
}
=/
=
(
Fi
,
E
,
F
}
)
with
Fi
=
F
,
IX
,
Y
,
Z
)
,
I
-4,23
=P
•
Y
=
F
,
+
F-
¥
+
1=3
div
É
0
•
É
=
1¥
,
J-y.f-zj.IE
.
%)
is
still
a
differential
operat or
a
scalar
function
4
=
JÉ+
'
+
¥-3
a
"
formal
inner
pr oduct
"
wanminliu@gmail.com
Theorem
1
.
The
divergence
as
flux
density
let
5
,
=
spher e
of
radius
e
with
center
P
SO
Vol
(F)
=
-34T
E3
Ñ
unit
outwar d
normal
vertov
field
on
Ig
É
smooth
thre e
dimensional
vector
field
.
Then
☒
E.
Ñds
dir
ÉCP
)
=
lim
sa
e→o+
Voll
Se
)
^
⇐
N
←
"
•
pye
7
→
41,0-1
L
wanminliu@gmail.com
#
3
Calculate
div
=P
and
cure
É
É
=
yT+zj+
✗
É
Sol
.
=p
=
(
y
,
Z
,
X
)
div
É
=p
.
É
=
¥9
-1¥
2-
+
I.
✗
=
0
,
i
i
E
cure
É=
Tx
É
=
!
¥
÷
/
y
z
✗
=
×
-
:*
.
-1¥
-
¥1
,
¥
-
¥4
=
(
-1
,
-1
,
-1
)
.
Or
we
write
cure
É=
-
I
-
J
-
E
☐
wanminliu@gmail.com
#
7
Calculate
div
=P
and
cure
É
É
=
fix
It
gly
,
I
+
hlz
)
É
Sol
'
div
I
=
÷
tix
,
-1%941+1-+414
=
fix
)
1-
gly
)
-1
h'
12-1
.
i
i
E
cure
É
=/
¥
I
,
E-
|
fix
941
hit
=
If
:-,
has
-3-+9×4+5
/
It
"
-
E.
has
)
+
III.
sin
-
Itu
)
=
8
☐
Remark
.
We
cannot
write
curl
É=o
Since
it
is
a
zero
vector
in
IR
?
it
is
not
the
zero
number
.
wanminliu@gmail.com
#
11
Calculate
dive
and
oureÉ
divÉ=
☐
•
É
=
3-xlwsoi-J-ys.no/t0--T--r--cosoI-stnoJ
=
-15in
¥
,
+
lwso
)
③
Sol
.
F-
(
wso
,
sine
.
°
)
¥=¥i¥¥=¥±j¥
-
Y=¥z④
with
0=0
IX.
Y
)
20
*
=
%=÷¥
.
_i=÷i⑤
tano
=
¥
.
So
0=0-1×7
)
=
arctan
(
¥-3
.
so
dwÉ=③-¥,y±
-
¥É+×÷y±;¥
,
É
Sino
=
¥
①
①
④
②
⑤
to
Y
✗
w•=
:#
②
=
,¥;¥=µ÷,÷ = ÷
.
with
r=i-YÑ
.
dayEI=¥.¥
.
wanminliu@gmail.com
t.ie
cure
É
=/
I
,
I
,
¥-4
Cost
Sino
0
=
f-
Sino
,
¥
will
,
¥
Sino
-
¥
coso
)
T
T
Sino
=
eoso
20
*
=¥ñ×÷
.
②
④
↳
wso
=
-
since
¥y=I÷ñ×¥
①
⑤
so
¥
Sino
-
¥
Gso
=
¥÷y→
+
¥÷%=
?
d-
we
cannot
write
it
as
curl
É=
0
.
curl
=
(
0
,
0
,
0
)
.
It
is
a
zero
vector
.
☐
wanminliu@gmail.com
§
17.2
Identities
µ
,
0×601=8
curl
grad
=3
If
É
is
conservativ e
,
i.e
-5=001
for
some
¢
Let
¢
,
4
be
smooth
scalar
fields
then
☐
✗
=
☐
xioos-o.ie#isirrotational
.
and
¥
,
I
be
smooth
vector
fields
lil
Px
*
E)
=
ply
.
E)
-
p2É
(
curl
curl
=
grad
div
-
Laplacian
)
(a)
71441 =474+4701
grad
4--84=1,1×4
,
4%-41
(b)
7.
*
E)
=
4)
•
Éto
•
E)
dive
=
☐
•
É=
+
+5¥
a
8×11051--1741*1=>+417×1 -7
(d)
V.
(-7×8)=6×1--7
.ci
-
E.
COXE
,
cure
É
=
☐
×
=
2
,
I
,
Iz
I.
÷
:L
(e)
8×1
✗
E)
=
6
.ci/E-ci.p--T
-
G.
E)
E-
(
F-
•
7)
É
Laplacian
operator
0--02=8.0
8201
=
p.hu/)--divgrad0
f-
I
☐
(
É
.ci/----xlPxaJ-cixl0xEi
+
(
E.
a)
Etta
.ME
=
lot
¢
+
¥ 01
(g)
p
.
(8×-7)=0
div
curl
=o
ÑÉ
=
(04-1,045,51--3)
.
CfÉ=TxÉ
.ie
.
E-
cure
E.
then
diva
:o)
(
G-
•
0
)É=G
,
+
GREY
-16,3¥
wanminliu@gmail.com
#
9
Let
Ñ=
✗
It
yjt
2-
É
computation
for
¥
(
ten
x
)
.
and
r
=
IF
/
.
Since
r
=
IF
/
=
(
4-
y
'
+
2-
2)
I
(a)
if
f-
is
a
differentiable
funtion
of
one
variable
,
show
that
so
2¥
=
-1,1×7-5+22
)
-
£
.
zx
=
¥
7.
(
fer
)
E)
=
rf.ir/t3fcr1
.
¥
ffinx
)
=
¥
Item
)
✗
+
firs
Us
(b)
Find
fin
=
firs
.
✗
1-
fin
if
first
is
solenoidal
for
r
-1-0
.
=
firs
.
¥
.
✗
+
fir
)
=
fin
+
tin
.
②
Proof
of
(a)
similarly
first
=
(
fer
)
×
,
ferry
,
firtz
)
¥
Itchy
)
=
fir
,
+
Hr
)
③
¥
Hint
)
=
tin
#
+
tin
.
④
g.
(
fer
)
F)
=
⇐
,
Ty
,
Cfcnx
,
fin
Y.tw
'
Z
)
go
put
②
③
④
into
①
=
¥14m
×
)
+
¥14m
Y
)
Hinz
?
①
g.
(
fer
)
F)
=
fir
)¥¥±
-131-14
=
fir
>
r
+
3
HH
.
☐
wanminliu@gmail.com
Recall
that
a
vector
field
É
is
so
called
solenoidal
in
a
domain
D
fir
)
=
-
f-
firs
fr
-1-0
)
if
div
É
=
o
in
D.
dt
In
=
-
f
f
(b)
if
fin
}
is
solenoidal
for
r
-1-0
.
If
f-
=
0
the
constant
0
function
,
then
then
div
(
first
)=0
in
D
,
g-
satisfies
the
above
equation
.
If
f
is
not
the
zero
function
,
we
then
hav e
with
the
domain
D=
Ird
1/10.01
}
d¥
=
-
Idr
That
is
f.
Him
F)
=
0
in
D.
but
=
-3
en
r
+
C
,
By
(a)
,
we
have
f-
=
Cr
-3
f-
'
(
r
)
r
+
3
fer
)
=
0
in
D
This
solution
is
so
fer
)
=
C
r
-3
missing
in
the
answ ers
of
the
textbook
,
or
fin
=
0
☐
&
then
first
=
10 .0.0
)
the
Inv entor
field
.
wanminliu@gmail.com
#
15
If
the
vector
fields
É
and
É
since
=p
and
E
are
conservative
,
are
smooth
and
conservative
.
show
we
can
write
¥
and
I
as
that
⇐
✗
É
is
solenoidal
.
É=
to
,
G-
=
04
Find
a
vector
potential
for
ÉXÉ
.
for
some
scalar
field
$
and
4
.
Assume
that
D
also
satisfies
the
assumption
we
compute
dirt
#
✗
E)
=p
.
(
Éx
E)
in
the
Theorem
5
.
Recall
a
vector
field
É
is
conservative
7.
(
É
✗
E)
=p
.
(6011×641)
if
it
is
of
the
form
¥
=
☐
at
☒
t.EE
"
"
-
"
"
"
YEE
"
for
some
scalar
field
4
.
Sol
.
We
show
that
ÉxE
is
solenoidal
,
=
0
To
find
a
vector
potential
,
we
need
i.
e.
div
(
É
✗
E)
=
0
in
D
,
to
recall
what
does
it
mean
,
where
D
is
the
common
domain
of
É
and
É
.
wanminliu@gmail.com
since
y
•
(
É
✗
E)
=
0
.
Theorem
5
.
Let
É
be
a
smooth
and
É
,
§
are
smooth
.
Solenoidal
vector
field
on
a
domain
D.
so
⇐
✗
E
is
a
smooth
solenoidal
Assume
the
domain
D
has
the
property
that
vector
field
in
D.
every
closed
surface
in
D
bounds
a
domain
Assume
that
D
also
satisfies
the
assumption
in
the
Theorem
5.
Then
by
Theorem
5
.
Contained
in
D
,
then
Éxci
=
☐
✗
(
ti
)
É
=
cure
E
f-
0
✗
E)
.
for
a
vector
field
Ñ
.
for
some
vector
field
É
defined
on
D.
we
wan t
to
find
such
IT
so
that
such
a
vector
tied
É
is
called
a
4)
✗
⑦
4)
=
✗
(
ti
)
vector
potential
of
the
vector
field
=P
.
The
idea
is
to
try
"
vector
formulas
"
latch
let
f)
X-D
Let
D=
1123140
,
o
,
on
the
IR
}
delete
the
In
particular
,
we
need
a
form
origin
,
then
such
D
does
NO T
satisfy
the
assumption
Creltor
field
)
✗
(
vector
field
)
in
the
Theorem
5
.
(C)
7×141=7
=
f)
✗
É
+
018×1=7
wanminliu@gmail.com
7×(0/04)
so
Pff
4)
001
)
4)
✗
4)
+417×10%41
=
(01-14)
✗
4)
so
Ñ=
004
is
a
vector
potential
☐
=
-
4)
✗
01011
Remark
.
The
vector
potential
is
NO T
unique
.
=
☐
(4)
✗
014 1
=
Éx
§
.
We
could
also
try
Hence
-47/0
is
also
a
vector
potential
.
8×148011
More
general
,
let
d
be
a
constant
14×1001+410%41
"
and
assume
0
EX
E
1
,
then
µ>=
>
$04
+
4-
a)
1-
4)
001
SO
8×140011=(014×64)
Recall
for
two
3-
dim
vectors
a-
and
B
is
also
a
vector
potential
.
a-
✗
bi
=
-
b-
✗
a-
.
wanminliu@gmail.com
We
want
to
find
a
vector
field
ti
#
17
show
that
so
that
☒
=
✗
e
"
I
+
ye2Zj
-
e2ZE
=P
=
ox
it
is
a
solenoidal
Vertov
field
.
Theorem
5
guarantee
the
existence
of
smh
Ñ
.
Find
a
vector
potential
for
it
.
Write
17=4-1
,
,
Ha
,
Hs
)
.
o×ñ=¥÷
-
¥
.
)i+f¥
.
-
¥×)i+f¥
-
¥1
:
Sol
.
We
want
to
show
so
we
want
to
Solve
(
Hi
,
Ha
.
Hs
)
for
equations
I
•
I
=
0
.
2¥
-
0¥
=
✗
e
"
①
É=
(
✗
e4
ye
"
,
-
e
"
)
{
1¥
-3¥
=
ye
"
②
8.
É
=
¥
(
✗
e
"
)
+
¥(
ye
"
)
+
¥1
-
e
't
)
3¥
-21¥
=
-
e
"
③
Try
③
:
let
H
,
=
ye
"
H
,
=
0
=
eZt
+
e
it
-
z
e
"
=
0
.
Then
①
2¥-
=
✗
EZZ
H
,
=
XYEZZ
-1/-142-1
So
É
is
a
smooth
solenoidal
vector
field
in
D=
IR
?
②
zye
"
-
ye
't
-1¥
=
ye
"
so
let
f-
=
f-
12-7
be
any
smooth
function
of
2-
.
Ñ=(
yea
,
0
,
✗
ye
"-
+
fits
)
is
a
vector
potenti al
.
☐
